If A and B are non - singular matrices of order `3xx3`, such that `A=(adjB)` and `B=(adjA)`, then det `(A)+det(B)` is equal to (where `det(M)` represents the determinant of matrix M and adj M represents the adjoint matrix of matrix M)

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we need to analyze the given conditions and use properties of determinants and adjoint matrices.

### Step 1: Understand the properties of adjoint matrices
We know that for any square matrix \( M \), the adjoint of \( M \), denoted as \( \text{adj}(M) \), has the property:
\[ M \cdot \text{adj}(M) = \det(M) I \]
where \( I \) is the identity matrix.

### Step 2: Apply the properties to the given matrices
Given that \( A = \text{adj}(B) \) and \( B = \text{adj}(A) \), we can substitute \( B \) in the first equation:
\[ A = \text{adj}(B) = \text{adj}(\text{adj}(A)) \]

### Step 3: Use the property of the adjoint of the adjoint
Using the property of adjoints:
\[ \text{adj}(\text{adj}(A)) = \det(A) A \]
Thus, we can write:
\[ A = \det(A) A \]

### Step 4: Rearranging the equation
From the equation \( A = \det(A) A \), we can rearrange it:
\[ A - \det(A) A = 0 \implies A(1 - \det(A)) = 0 \]
Since \( A \) is non-singular (det(A) ≠ 0), we can conclude:
\[ 1 - \det(A) = 0 \implies \det(A) = 1 \]

### Step 5: Find the determinant of \( B \)
Now, using the relation \( B = \text{adj}(A) \), we can apply the same property:
\[ B = \text{adj}(A) = \det(A) A = 1 \cdot A = A \]
Thus, we find:
\[ \det(B) = \det(A) = 1 \]

### Step 6: Calculate \( \det(A) + \det(B) \)
Now we can find:
\[ \det(A) + \det(B) = 1 + 1 = 2 \]

### Conclusion
Therefore, the final answer is:
\[ \det(A) + \det(B) = 2 \]